Question

A block of mass 3 $kg$ is released from the rest at top of the inclined plane. If length and height of the inclined plane is 6 m and 4 m respectively. Calculate its speed at the bottom of inclined plane. Assume the force of friction 16 N on the block is a constant.

• A. 2 m/s
• B. 3 m/s
• C. 4 m/s
• D. 5 m/s
• E. 6 m/s

Answer 1

Answer: (C)

4 m/s

Acceleration of the block along the inclined surface=$a=gsin\theta -\frac{f}{m}$

$=10m/s^2\times \frac{4}{6}-\frac{16N}{3kg}$

$=\frac{4}{3}m/s^2$

Distance traveled along the surface=$s=6m$

Thus speed attained by the block=$\sqrt{2as}=4m/s$